Tutorial 4: Regression in R

1. Linear Regression Models

1.1 Concept heads up

Linear Regression method is one of the most common research methods examining the linear relationship of the dependent variable \(Y\) and independent variable(s) \(X\).

If you want to determine the value of a continuous response variable \(Y\) given the values of \(p >= 1\) independent explanatory variables \(X_1, X_2, . . ., X_p\). The overarching model is defined as

\[Y = \beta_0+\beta_1X_1+\beta_2X_2+\cdots+\beta_pX_p+\epsilon,\] where \(\beta_0,\cdots,\beta_p\) are the regression coefficients and we assume indepent, normally distributed residuals \(\epsilon \sim N(0,\sigma)\) around the mean.

The model to be fitted is given in terms of the mean response, conditional upon a particular realization of the set of explanatory variables \[\hat{y} = E[Y|X_1=x_1,X_2=x_2,\cdots,X_p=x_p]=\hat{\beta}_0+\hat{\beta_1}x_1+\cdots+\hat{\beta}_px_p,\] where \(\hat{\beta}_j\) represent estimates of the regression coefficients.

For the \(n\) data records, the \(\hat{\beta}_j\) are found as the values that minimize the sum

\[\sum^n_{i=1}[y_i-(\hat{\beta}_0+\hat{\beta_1}x_{1,i}+\cdots+\hat{\beta}_px_{p,i})]^2,\] where \(x_{j,i}\) is the observed value of individual \(i\) for explanatory variable \(X_j\) and \(y_i\) is their response value.

The computations involved in minimizing this squared distance are made much easier by a matrix representation of the data. When dealing with \(n\) multivariate observations, you can write the linear model as \[\textbf{Y}=\textbf{X}\cdot\boldsymbol{\beta}+\boldsymbol{\epsilon},\]

where \(\boldsymbol{Y}\) and \(\boldsymbol{\epsilon}\) denote \(n\times 1\) column matrices such that

\[ \mathbf{Y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} \quad \text{and} \quad \boldsymbol{\epsilon} = \begin{bmatrix} \epsilon_1 \\ \epsilon_2 \\ \vdots \\ \epsilon_n \end{bmatrix}. \] Here, \(y_i\) and \(\epsilon_i\) refer to the response observation and random error term for the \(i\)-th individual. The quantity \(\boldsymbol{\beta}\) is a \((p + 1) \times 1\) column matrix of the regression coefficients, and the observed predictor data for all individuals and explanatory variables are stored in an \(n \times (p + 1)\) matrix \(\mathbf{X}\), called the design matrix:

\[ \boldsymbol{\beta} = \begin{bmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p \end{bmatrix} \quad \text{and} \quad \mathbf{X} = \begin{bmatrix} 1 & x_{1,1} & \cdots & x_{1,p} \\ 1 & x_{2,1} & \cdots & x_{2,p} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n,1} & \cdots & x_{n,p} \end{bmatrix}. \] The minimization of least square providing the estimated regression coefficient values is then found with the following calculation: \[ \hat{\boldsymbol{\beta}} = \begin{bmatrix} \hat{\beta}_0 \\ \hat{\beta}_1 \\ \vdots \\ \hat{\beta}_p \end{bmatrix} = \left( \mathbf{X}^{\top} \mathbf{X} \right)^{-1} \mathbf{X}^{\top} \mathbf{Y} \]

1.2 Simple Linear regression models

An easy example using the equation we just reviewed

You can manually estimate the \(\beta_j,(j=0,1,\cdots,p)\) in R. As a quick demonstration, let’s say you have two predictor variables: X1 as continuous and X2 as binary. Your target regression equation is therefore \(\hat{y}=\hat{\beta}_0+\hat{\beta}_1x_1+\hat{\beta}_2x_2\).

Suppose you collect the following data, where the response data, data for \(X_1\), and data for \(X_2\), for \(n = 8\) individuals, are given in the columns \(y\), \(x_1\), and \(x_2\), respectively.

demo.data <- data.frame(y=c(1.55,0.42,1.29,0.73,0.76,-1.09,1.41,-0.32),
                           x1=c(1.13,-0.73,0.12,0.52,-0.54,-1.15,0.20,-1.09),
                            x2=c(1,0,1,1,0,1,0,1))
demo.data

##       y    x1 x2
## 1  1.55  1.13  1
## 2  0.42 -0.73  0
## 3  1.29  0.12  1
## 4  0.73  0.52  1
## 5  0.76 -0.54  0
## 6 -1.09 -1.15  1
## 7  1.41  0.20  0
## 8 -0.32 -1.09  1

Question: Now, try to compute the point estimates in \(\boldsymbol{\beta}\) for the linear model using the equation we deducted earlier.

Example using function lm()

Let’s first get sample data.

#install.packages(“car”) ##if this is a new package we need to install first.

library(car) #We have to library the package every time we use

We are using the dataset “Prestige” * in the car package for this guide. We can use the help command to access the codebook:

help("Prestige")
View(Prestige)

Question: Can you use function lm() to describe the relationship between the dependent variable(prestige) and one independent variable(education)? And how can we interpret the model?

Hint: you can use summary to view the result.

Log transformation:

It’s a useful practice to transform the variable to it’s log form when doing the regression. You can either transform the variable beforehand or do so in the equation.

Note that in some cases the variable contains the value of 0 so we cannot directly transform them to the log-form. We can add the value of 1 to every observation and then do the transformation.

Question: Use the same data, now, use log transformaton on income, and do another linear model regarding the relationship between the response variable ‘prestige’ and the three independent variables ‘education’, ‘income’ and ‘women’. And how can we interpret the model now?

Predicted values and Residuals:

Question Using regression model 2, how can you predict the value of prestige and how do you obtain the residuals?

Regression with Categorical variables

Categorical variables play a big part in the linear regression too. Note that we only talk about where independent variables are categorical variables in this guide. When the dependent variable is a categorical variable, you may consider the alternatives of linear regression like logit regression and multinomial regression. In this dataset, type is a categorical variable that describes the type of occupation.

Question Construct a linear model that describe the relationship between the response variable ‘prestige’ and three independent variables ‘education’, ‘log(income)’ and ‘type’.

Categorical variables with interaction terms

Sometimes we are interested in the interaction terms of two variables. Here is an example of interaction term with categorical variables.

Question We would like to use the same independent variables as before, but now, we want to include the interation between ‘type’ and ‘education’, ‘type’ and ‘log(income)’. Can you consider how to construct this linear model?

2.Assumption Diagnostics and Regression Trouble Shooting

It’s a good habit to check for the assumptions of linear regressions to see whether the model is good. We will use the raw variables (without transforming) to construct a model and then we can test for the assumptions. Note that these commands, if not specifically stated, are from the ‘car’ package.

2.1 Normality Check

It’s important to check whether the normality assumption holds for the model.

Now we consider the linear model

model1 <- lm(prestige ~ education + income + type, data = Prestige)

Question How can we test the normality assumption? Hint: you can use the matrial from tutorial 3: ANOVA to check the normality, both from qqPlot() and `shapiro.test()

2.2 Heteroskedasticity Check

We can use the ncvTest() or leveneTest(), etc. to test for equal variances. We can also consult with the residual plots, which we will discuss later, whether the model meets the homoscedasticity assumption.

Question Please use ncvTest() on model1 and check the result.

2.3 Multicollinearity Check

We want to know whether we have too many variables that have high correlation with each other.

“When there are strong linear relationships among the predictors in a regression analysis, the precision of the estimated regression coefficients in linear models declines compared to what it would have been were the predictors uncorrelated with each other” (Fox:359)

Question Please use vif() on model1 and check the result. A GVIF > 4 suggests collinearity.

2.4 Residual Plots

We can also plot the residual plots on model1.

Question Please plot the residuals plots and comment.

2.5 Outliers

Outlier Identification is also significant.

We can first look at the QQplot.

model1 <- lm(prestige ~ education + income + type, data = Prestige)
qqPlot(model1,id.n=2)

## medical.technicians  electronic.workers 
##                  31                  82

2.6 Influential/High leverage points

We can look at the Influential/ high leverage points.

#Influence Plots
influencePlot(model1)

##                        StudRes        Hat       CookD
## general.managers    -1.3134574 0.33504477 0.172503975
## physicians          -0.3953204 0.22420309 0.009115491
## medical.technicians  2.8210910 0.06858836 0.109052582
## electronic.workers   2.2251940 0.02701237 0.026372394

Example 1: Factors in linear regression

Interpreting coefficients of factor variables

In the case of quantitative predictors, we’re more or less comfortable with the interpretation of the linear model coefficient as a “slope” or a “unit increase in outcome per unit increase in the covariate”. This isn’t the right interpretation for factor variables. In particular, the notion of a slope or unit change no longer makes sense when talking about a categorical variable. E.g., what does it even mean to say “unit increase in major” when studying the effect of college major on future earnings?

To understand what the coefficients really mean, let’s go to the birthwt data and try regressing birthweight on mother’s race and mother’s age.

library(tidyverse)
# Load data from MASS into a tibble
# Rename variables
# Recode categorical variables
birthwt <- as_tibble(MASS::birthwt) %>%
  rename(birthwt.below.2500 = low, 
         mother.age = age,
         mother.weight = lwt,
         mother.smokes = smoke,
         previous.prem.labor = ptl,
         hypertension = ht,
         uterine.irr = ui,
         physician.visits = ftv,
         birthwt.grams = bwt)  %>%
  mutate(race = recode_factor(race, `1` = "white", `2` = "black", `3` = "other")) %>%
  mutate_at(c("mother.smokes", "hypertension", "uterine.irr", "birthwt.below.2500"),
            ~ recode_factor(.x, `0` = "no", `1` = "yes"))

# Fit regression model
birthwt.lm <- lm(birthwt.grams ~ race + mother.age, data = birthwt)

# Regression model summary
summary(birthwt.lm)

## 
## Call:
## lm(formula = birthwt.grams ~ race + mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2131.57  -488.02    -1.16   521.87  1757.07 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2949.979    255.352  11.553   <2e-16 ***
## raceblack   -365.715    160.636  -2.277   0.0240 *  
## raceother   -285.466    115.531  -2.471   0.0144 *  
## mother.age     6.288     10.073   0.624   0.5332    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 715.7 on 185 degrees of freedom
## Multiple R-squared:  0.05217,    Adjusted R-squared:  0.0368 
## F-statistic: 3.394 on 3 and 185 DF,  p-value: 0.01909

Note that there are two coefficients estimated for the race variable (raceother and racewhite). What’s happening here?

When you put a factor variable into a regression, you’re allowing a different intercept at every level of the factor. In the present example, you’re saying that you want to model birthwt.grams as

Baby’s birthweight = Intercept(based on mother’s race) + β * mother’s age

We can rewrite this more succinctly as: \[ y=\mathrm{Intercept}_{race}+\beta\times age \] Essentially you’re saying that your data is broken down into 3 racial groups, and you want to model your data as having the same slope governing how birthweight changes with mother’s age, but potentially different intercepts. Here’s a picture of what’s happening.

# Calculate race-specific intercepts
intercepts <- c(coef(birthwt.lm)["(Intercept)"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceblack"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceother"])

lines.df <- data.frame(intercepts = intercepts,
                       slopes = rep(coef(birthwt.lm)["mother.age"], 3),
                       race = levels(birthwt$race))

ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +
  geom_point() +
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)

How do we interpret the 2 race coefficients? For categorical variables, the interpretation is relative to the given baseline. The baseline is just whatever level comes first (here, “white”). E.g., the estimate of raceother means that the estimated intercept is -285.47 higher among “other” race mothers compared to black mothers. Similarly, the estimated intercept is -365.72 higher for black mothers than white mothers.

Another way of putting it: Among mothers of the same age, babies of black mothers are born on average weighing -365.72g more than babies of white mothers.

Why is one of the levels missing in the regression?

As you’ve already noticed, there is no coefficient called “racewhite” in the estimated model. This is because this coefficient gets absorbed into the overall (Intercept) term.

Let’s peek under the hood. Using the model.matrix() function on our linear model object, we can get the data matrix that underlies our regression. Here are the first 20 rows.

head(model.matrix(birthwt.lm), 20)

##    (Intercept) raceblack raceother mother.age
## 1            1         1         0         19
## 2            1         0         1         33
## 3            1         0         0         20
## 4            1         0         0         21
## 5            1         0         0         18
## 6            1         0         1         21
## 7            1         0         0         22
## 8            1         0         1         17
## 9            1         0         0         29
## 10           1         0         0         26
## 11           1         0         1         19
## 12           1         0         1         19
## 13           1         0         1         22
## 14           1         0         1         30
## 15           1         0         0         18
## 16           1         0         0         18
## 17           1         1         0         15
## 18           1         0         0         25
## 19           1         0         1         20
## 20           1         0         0         28

Even though we think of the regression birthwt.grams ~ race + mother.age as being a regression on two variables (and an intercept), it’s actually a regression on 3 variables (and an intercept). This is because the race variable gets represented as two dummy variables: one for race == other and the other for race == black.

Why isn’t there a column for representing the indicator of race == white? This gets back to our colinearity issue. By definition, we have that

racewhite + raceother + raceblack = 1 = (Intercept)

This is because for every observation, one and only one of the race dummy variables will equal 1. Thus the group of 4 variables {racewhite, raceother, raceblack, (Intercept)} is perfectly colinear, and we can’t include all 4 of them in the model. The default behavior in R is to remove the dummy corresponding to the first level of the factor (here, racewhite), and to keep the rest.

Interaction terms

Let’s go back to the regression line plot we generated above.

ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +
  geom_point() +
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)

We can have similar plots by using the geom_smooth or stat_smooth commands in ggplot. Compare the plot above to the following.

ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +
  geom_point() +
  stat_smooth(method = "lm", se = FALSE, fullrange = TRUE)

In this case we have not only race-specific intercepts, but also race-specific slopes. The plot above corresponds to the model:

Baby’s birthweight = Intercept(based on mother’s race) + β(based on mother’s race) * mother’s age

We can rewrite this more succinctly as: \[ y=\mathrm{Intercept}_{race}+\beta_{race}\times age \] To specify this interaction model in R, we use the following syntax

birthwt.lm.interact <- lm(birthwt.grams ~ race * mother.age, data = birthwt)
summary(birthwt.lm.interact)

## 
## Call:
## lm(formula = birthwt.grams ~ race * mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2182.35  -474.23    13.48   523.86  1496.51 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)           2583.54     321.52   8.035 1.11e-13 ***
## raceblack             1022.79     694.21   1.473   0.1424    
## raceother              326.05     545.30   0.598   0.5506    
## mother.age              21.37      12.89   1.658   0.0991 .  
## raceblack:mother.age   -62.54      30.67  -2.039   0.0429 *  
## raceother:mother.age   -26.03      23.20  -1.122   0.2633    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 710.7 on 183 degrees of freedom
## Multiple R-squared:  0.07541,    Adjusted R-squared:  0.05015 
## F-statistic: 2.985 on 5 and 183 DF,  p-value: 0.01291

We now have new terms appearing. Terms like raceblack:mother.age are deviations from the baseline slope (the coefficient of mother.age in the model) in the same way that terms like raceblack are deviations from the baseline intercept. This models says that:

On average among white mothers, every additional year of age is associated with a 21.37g increase in the birthweight of the baby.

To get the slope for black mothers, we need to add the interaction term to the baseline. \[ \beta_{\text{raceblack}}=\beta_{\text{racewhite}}+\beta_{\text{raceblack:mother.age}}=\text{mother.age}+\text{raceblack:mother.age}=21.3727574+(-62.5379308)= -41.1651733 \] This slope estimate is negative, which agrees with the regression plot above.

Is a categorical variable in a regression statistically significant?

Last class we considered modelling birthweight as a linear function of mother’s age, allowing for a race-specific intercept for each of the three race categories. This model is fit again below.

birthwt.lm <- lm(birthwt.grams ~ race + mother.age, data = birthwt)

Here’s a visualization of the model fit that we wound up with. Note that while there are 3 lines shown, this is a visualization of just one model: birthwt.grams ~ race + mother.age. This model produces 3 lines because the coefficients of the race variable result in different intercepts.

# Calculate race-specific intercepts
intercepts <- c(coef(birthwt.lm)["(Intercept)"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceblack"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceother"])

lines.df <- data.frame(intercepts = intercepts,
                       slopes = rep(coef(birthwt.lm)["mother.age"], 3),
                       race = levels(birthwt$race))

ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +   
  geom_point() +
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)

At this stage we may be interested in assessing whether the race variable is statistically significant. i.e., Does including the race variable significantly improve the fit of our model, or is the simpler model birthwt.grams ~ mother.age just as good?

Essentially, we want to know if the race-specific intercepts capture significantly more variation in the outcome (birthweight) than the single intercept model, or if allowing for different intercepts isn’t doing much more than capturing random fluctuations in the data.

Here’s a picture of the two models we’re comparing:

library(gridExtra)
plot.complex <- ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +
  geom_point() +
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)

# Single intercept model (birthwt.grams ~ mother.age)
p <- ggplot(birthwt, aes(x = mother.age, y = birthwt.grams))
plot.simple <- p + geom_point(aes(colour = race)) + stat_smooth(method = "lm")

grid.arrange(plot.complex, plot.simple, ncol = 2)

To test this hypothesis, we use the anova function (not to be confused with the aov function). This function compares two nested models, accounting for their residual sums of squares (how well they fit the data) and their complexity (how many more variables are in the larger model) to assess statistical significance.

# Fit the simpler model with mother.age as the only predictor
birthwt.lm.simple <- lm(birthwt.grams ~ mother.age, data = birthwt)

# Compare to more complex model
anova(birthwt.lm.simple, birthwt.lm)

## Analysis of Variance Table
## 
## Model 1: birthwt.grams ~ mother.age
## Model 2: birthwt.grams ~ race + mother.age
##   Res.Df      RSS Df Sum of Sq      F  Pr(>F)  
## 1    187 99154173                              
## 2    185 94754346  2   4399826 4.2951 0.01502 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

This output tells us that the race variable is statistically significant: It is unlikely that the improvement in fit when the add the race variable is simply due to random fluctuations in the data. Thus it is important to consider race when modeling how birthweight depends on the mother’s age.

Is an interaction term significant?

Assessing significance of interaction terms operates on the same principle. We once again ask whether the improvement in model fit is worth the increased complexity of our model. For instance, consider the example we saw last class, where we allowed for a race-specific slope in addition to the race-specific intercept from before.

birthwt.lm.interact <- lm(birthwt.grams ~ race * mother.age, data = birthwt)
summary(birthwt.lm.interact)

## 
## Call:
## lm(formula = birthwt.grams ~ race * mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2182.35  -474.23    13.48   523.86  1496.51 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)           2583.54     321.52   8.035 1.11e-13 ***
## raceblack             1022.79     694.21   1.473   0.1424    
## raceother              326.05     545.30   0.598   0.5506    
## mother.age              21.37      12.89   1.658   0.0991 .  
## raceblack:mother.age   -62.54      30.67  -2.039   0.0429 *  
## raceother:mother.age   -26.03      23.20  -1.122   0.2633    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 710.7 on 183 degrees of freedom
## Multiple R-squared:  0.07541,    Adjusted R-squared:  0.05015 
## F-statistic: 2.985 on 5 and 183 DF,  p-value: 0.01291

Here’s a side-by-side visual comparison of therace + mother.age model and the race + mother.age + race*mother.age interaction model.

plot.interact <- ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +   
  geom_point() +
  stat_smooth(method = "lm", se = FALSE, fullrange = TRUE)
grid.arrange(plot.interact, plot.complex, ncol = 2)

So, do the lines with different slopes fit the data significantly better than the common slope model? Let’s compare the two with the anova() function.

anova(birthwt.lm, birthwt.lm.interact)

## Analysis of Variance Table
## 
## Model 1: birthwt.grams ~ race + mother.age
## Model 2: birthwt.grams ~ race * mother.age
##   Res.Df      RSS Df Sum of Sq      F Pr(>F)
## 1    185 94754346                           
## 2    183 92431148  2   2323199 2.2998 0.1032

This \(p\)-value turns out to not be statistically significant. So even though the estimated slopes in the interaction model look very different, our estimates are quite variable, so we don’t have enough evidence to conclude that the interaction term (different slopes) is providing significant additional explanatory power over the simpler race + mother.age model.

Is my complex model signficantly better than a simpler one?

The testing strategy above applies to any two nested models. Here’s an example where we add in a few more variables and see how it compares to the race + mother.age model from earlier.

birthwt.lm.complex <- lm(birthwt.grams ~ mother.smokes + physician.visits + race + mother.age, data = birthwt)

summary(birthwt.lm.complex)

## 
## Call:
## lm(formula = birthwt.grams ~ mother.smokes + physician.visits + 
##     race + mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2335.06  -455.16    31.74   499.29  1623.57 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)      3282.407    261.326  12.561  < 2e-16 ***
## mother.smokesyes -424.651    110.371  -3.847 0.000165 ***
## physician.visits   14.391     48.953   0.294 0.769102    
## raceblack        -444.340    156.586  -2.838 0.005057 ** 
## raceother        -445.161    119.666  -3.720 0.000265 ***
## mother.age          1.547      9.996   0.155 0.877155    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 691.7 on 183 degrees of freedom
## Multiple R-squared:  0.1241, Adjusted R-squared:  0.1001 
## F-statistic: 5.184 on 5 and 183 DF,  p-value: 0.000179

Let’s compare to our earlier model:

anova(birthwt.lm, birthwt.lm.complex)

## Analysis of Variance Table
## 
## Model 1: birthwt.grams ~ race + mother.age
## Model 2: birthwt.grams ~ mother.smokes + physician.visits + race + mother.age
##   Res.Df      RSS Df Sum of Sq      F    Pr(>F)    
## 1    185 94754346                                  
## 2    183 87567280  2   7187067 7.5098 0.0007336 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Highly significant! This is probably due to the fact that mother’s smoking status has a tremendously high association with birthweight.

Example 2: Summarizing stratified regressions

Gapminder life expectancy data

(gapminder <- read_delim("http://www.andrew.cmu.edu/user/achoulde/94842/data/gapminder_five_year.txt",
                        delim = "\t")) # Load data

## # A tibble: 1,704 × 6
##    country      year      pop continent lifeExp gdpPercap
##    <chr>       <dbl>    <dbl> <chr>       <dbl>     <dbl>
##  1 Afghanistan  1952  8425333 Asia         28.8      779.
##  2 Afghanistan  1957  9240934 Asia         30.3      821.
##  3 Afghanistan  1962 10267083 Asia         32.0      853.
##  4 Afghanistan  1967 11537966 Asia         34.0      836.
##  5 Afghanistan  1972 13079460 Asia         36.1      740.
##  6 Afghanistan  1977 14880372 Asia         38.4      786.
##  7 Afghanistan  1982 12881816 Asia         39.9      978.
##  8 Afghanistan  1987 13867957 Asia         40.8      852.
##  9 Afghanistan  1992 16317921 Asia         41.7      649.
## 10 Afghanistan  1997 22227415 Asia         41.8      635.
## # ℹ 1,694 more rows

Maximum life expectancy for each continent

Before diving into a regression, let’s look at some data summaries that use some familiar functions. Here’s how we can form a table showing the maximum life expectancy on each continent each year, and the country that attained that maximum.

gapminder %>%
  group_by(continent, year) %>%
  summarize(max.life.exp = max(lifeExp),
            country = country[which.max(lifeExp)])

## # A tibble: 60 × 4
## # Groups:   continent [5]
##    continent  year max.life.exp country  
##    <chr>     <dbl>        <dbl> <chr>    
##  1 Africa     1952         52.7 Reunion  
##  2 Africa     1957         58.1 Mauritius
##  3 Africa     1962         60.2 Mauritius
##  4 Africa     1967         61.6 Mauritius
##  5 Africa     1972         64.3 Reunion  
##  6 Africa     1977         67.1 Reunion  
##  7 Africa     1982         69.9 Reunion  
##  8 Africa     1987         71.9 Reunion  
##  9 Africa     1992         73.6 Reunion  
## 10 Africa     1997         74.8 Reunion  
## # ℹ 50 more rows

Fitting a linear model for each country

We’re now going to go through an example where we get the life expectancy in 1952 and the rate of change in life expectancy over time for each country. The rate of change will be obtained by regressing lifeExp on year.

(1) Figure out how to extract the desired information.

Let’s start with the data for a single country.

country.name <- "Ireland"  # Pick a country
gapminder.sub <- filter(gapminder, country == country.name)  # Pull data for this country
gapminder.sub

## # A tibble: 12 × 6
##    country  year     pop continent lifeExp gdpPercap
##    <chr>   <dbl>   <dbl> <chr>       <dbl>     <dbl>
##  1 Ireland  1952 2952156 Europe       66.9     5210.
##  2 Ireland  1957 2878220 Europe       68.9     5599.
##  3 Ireland  1962 2830000 Europe       70.3     6632.
##  4 Ireland  1967 2900100 Europe       71.1     7656.
##  5 Ireland  1972 3024400 Europe       71.3     9531.
##  6 Ireland  1977 3271900 Europe       72.0    11151.
##  7 Ireland  1982 3480000 Europe       73.1    12618.
##  8 Ireland  1987 3539900 Europe       74.4    13873.
##  9 Ireland  1992 3557761 Europe       75.5    17559.
## 10 Ireland  1997 3667233 Europe       76.1    24522.
## 11 Ireland  2002 3879155 Europe       77.8    34077.
## 12 Ireland  2007 4109086 Europe       78.9    40676.

# Scatterplot of life exp vs year
# with a regression line overlaid
ggplot(data = gapminder.sub, mapping = aes(x = year, y = lifeExp)) +
  geom_point() +
  geom_smooth(method = "lm") +
  ggtitle(paste("Life expectancy in", country.name))

We can confirm that it’s a pretty good model for other countries as well, though not for all of them

ggplot(data = gapminder, aes(x = year, y = lifeExp)) +
  facet_wrap( ~ country) +
  geom_point() +
  geom_smooth(method = "lm")

Now let’s fit a regression and extract the slope.

life.exp.lm <- lm(lifeExp ~ year, data = gapminder.sub) # Fit model
coef(life.exp.lm) # Get coefficients

##  (Intercept)         year 
## -321.1399594    0.1991196

coef(life.exp.lm)["year"]  # The slope that we wanted

##      year 
## 0.1991196

coef(lm(lifeExp ~ year, data = gapminder.sub))["year"]

##      year 
## 0.1991196

(2) Extract information for each country

Here we’ll extract the slope in the way we practiced above, and we’ll also extract the “origin” life expectancy: the given country’s life expectancy in 1952, the first year of our data. For the purpose of plotting we’ll also want the continent information, so we’ll capture that in this call too.

progress.df <- gapminder %>%
  group_by(country) %>%
  summarize(continent = continent[1],
            origin = lifeExp[year == 1952],
            slope =  lm(lifeExp ~ year)$coef["year"]) 

progress.df

## # A tibble: 142 × 4
##    country     continent origin slope
##    <chr>       <chr>      <dbl> <dbl>
##  1 Afghanistan Asia        28.8 0.275
##  2 Albania     Europe      55.2 0.335
##  3 Algeria     Africa      43.1 0.569
##  4 Angola      Africa      30.0 0.209
##  5 Argentina   Americas    62.5 0.232
##  6 Australia   Oceania     69.1 0.228
##  7 Austria     Europe      66.8 0.242
##  8 Bahrain     Asia        50.9 0.468
##  9 Bangladesh  Asia        37.5 0.498
## 10 Belgium     Europe      68   0.209
## # ℹ 132 more rows

What can we learn from this output?

Let’s summarize our findings by creating a bar chart for the origin and slope, with the bars colored by continent. We’ll do this in ggplot.

Plotting origins coloured by continent

# Reorder country factor by origin
# Construct bar chart
progress.df %>%
  mutate(country = reorder(country, origin)) %>%
  ggplot(aes(x = country, y = origin, fill = continent)) +
  geom_bar(stat = "identity") +
  theme(axis.text.x = element_text(angle = 60, vjust = 1, hjust = 1)) 

Plotting slopes coloured by continent

# Reorder country factor by slope
# Construct bar chart
progress.df %>%
  mutate(country = reorder(country, slope)) %>%
  ggplot(aes(x = country, y = slope, fill = continent)) +
  geom_bar(stat = "identity") +
  theme(axis.text.x = element_text(angle = 60, vjust = 1, hjust = 1)) 

These are very interesting plots. What can you tell from looking at them?

Looking at per capita GDP by year

Let’s start by looking at some plots of how GDP per capita varied by year

# qplot(year, gdpPercap, facets = ~ country, data = gapminder, colour = continent) +
ggplot(data = gapminder, mapping = aes(x = year, y = gdpPercap, colour = continent)) +
  geom_point() +
  facet_wrap(~ country, ncol=12) +
  theme(axis.text.x = element_text(angle = 90, hjust = 1))

What if we want to rearrange the plots by continent? This can be done by changing the order of the country level.

# First step: reorder the countries by continent
# Produce a data frame that has the countries ordered alphabetically within continent
# Arrange sorts the data according to the variable(s) provided
country.df <- gapminder %>%
  group_by(country) %>%
  summarize(continent = continent[1]) %>%
  arrange(continent)

gapminder.ordered <- gapminder %>%
  mutate(country = factor(country, levels = country.df$country))

# Let's make sure that things are now ordered correctly...
levels(gapminder.ordered$country)

##   [1] "Algeria"                  "Angola"                  
##   [3] "Benin"                    "Botswana"                
##   [5] "Burkina Faso"             "Burundi"                 
##   [7] "Cameroon"                 "Central African Republic"
##   [9] "Chad"                     "Comoros"                 
##  [11] "Congo, Dem. Rep."         "Congo, Rep."             
##  [13] "Cote d'Ivoire"            "Djibouti"                
##  [15] "Egypt"                    "Equatorial Guinea"       
##  [17] "Eritrea"                  "Ethiopia"                
##  [19] "Gabon"                    "Gambia"                  
##  [21] "Ghana"                    "Guinea"                  
##  [23] "Guinea-Bissau"            "Kenya"                   
##  [25] "Lesotho"                  "Liberia"                 
##  [27] "Libya"                    "Madagascar"              
##  [29] "Malawi"                   "Mali"                    
##  [31] "Mauritania"               "Mauritius"               
##  [33] "Morocco"                  "Mozambique"              
##  [35] "Namibia"                  "Niger"                   
##  [37] "Nigeria"                  "Reunion"                 
##  [39] "Rwanda"                   "Sao Tome and Principe"   
##  [41] "Senegal"                  "Sierra Leone"            
##  [43] "Somalia"                  "South Africa"            
##  [45] "Sudan"                    "Swaziland"               
##  [47] "Tanzania"                 "Togo"                    
##  [49] "Tunisia"                  "Uganda"                  
##  [51] "Zambia"                   "Zimbabwe"                
##  [53] "Argentina"                "Bolivia"                 
##  [55] "Brazil"                   "Canada"                  
##  [57] "Chile"                    "Colombia"                
##  [59] "Costa Rica"               "Cuba"                    
##  [61] "Dominican Republic"       "Ecuador"                 
##  [63] "El Salvador"              "Guatemala"               
##  [65] "Haiti"                    "Honduras"                
##  [67] "Jamaica"                  "Mexico"                  
##  [69] "Nicaragua"                "Panama"                  
##  [71] "Paraguay"                 "Peru"                    
##  [73] "Puerto Rico"              "Trinidad and Tobago"     
##  [75] "United States"            "Uruguay"                 
##  [77] "Venezuela"                "Afghanistan"             
##  [79] "Bahrain"                  "Bangladesh"              
##  [81] "Cambodia"                 "China"                   
##  [83] "Hong Kong, China"         "India"                   
##  [85] "Indonesia"                "Iran"                    
##  [87] "Iraq"                     "Israel"                  
##  [89] "Japan"                    "Jordan"                  
##  [91] "Korea, Dem. Rep."         "Korea, Rep."             
##  [93] "Kuwait"                   "Lebanon"                 
##  [95] "Malaysia"                 "Mongolia"                
##  [97] "Myanmar"                  "Nepal"                   
##  [99] "Oman"                     "Pakistan"                
## [101] "Philippines"              "Saudi Arabia"            
## [103] "Singapore"                "Sri Lanka"               
## [105] "Syria"                    "Taiwan"                  
## [107] "Thailand"                 "Vietnam"                 
## [109] "West Bank and Gaza"       "Yemen, Rep."             
## [111] "Albania"                  "Austria"                 
## [113] "Belgium"                  "Bosnia and Herzegovina"  
## [115] "Bulgaria"                 "Croatia"                 
## [117] "Czech Republic"           "Denmark"                 
## [119] "Finland"                  "France"                  
## [121] "Germany"                  "Greece"                  
## [123] "Hungary"                  "Iceland"                 
## [125] "Ireland"                  "Italy"                   
## [127] "Montenegro"               "Netherlands"             
## [129] "Norway"                   "Poland"                  
## [131] "Portugal"                 "Romania"                 
## [133] "Serbia"                   "Slovak Republic"         
## [135] "Slovenia"                 "Spain"                   
## [137] "Sweden"                   "Switzerland"             
## [139] "Turkey"                   "United Kingdom"          
## [141] "Australia"                "New Zealand"

ggplot(data = gapminder.ordered, mapping = aes(x = year, y = gdpPercap, colour = continent)) +
  geom_point() +
  facet_wrap(~ country, ncol=12) +
  theme(axis.text.x = element_text(angle = 90, hjust = 1)) + 
  stat_smooth(method = "lm") 

In the case of quantitative predictors, we’re more or less comfortable with the interpretation of the linear model coefficient as a “slope” or a “unit increase in outcome per unit increase in the covariate”. This isn’t the right interpretation for factor variables. In particular, the notion of a slope or unit change no longer makes sense when talking about a categorical variable. E.g., what does it even mean to say “unit increase in major” when studying the effect of college major on future earnings?

To understand what the coefficients really mean, let’s go to the birthwt data and try regressing birthweight on mother’s race and mother’s age.

library(tidyverse)
# Load data from MASS into a tibble
# Rename variables
# Recode categorical variables
birthwt <- as_tibble(MASS::birthwt) %>%
  rename(birthwt.below.2500 = low, 
         mother.age = age,
         mother.weight = lwt,
         mother.smokes = smoke,
         previous.prem.labor = ptl,
         hypertension = ht,
         uterine.irr = ui,
         physician.visits = ftv,
         birthwt.grams = bwt)  %>%
  mutate(race = recode_factor(race, `1` = "white", `2` = "black", `3` = "other")) %>%
  mutate_at(c("mother.smokes", "hypertension", "uterine.irr", "birthwt.below.2500"),
            ~ recode_factor(.x, `0` = "no", `1` = "yes"))

# Fit regression model
birthwt.lm <- lm(birthwt.grams ~ race + mother.age, data = birthwt)

# Regression model summary
summary(birthwt.lm)

## 
## Call:
## lm(formula = birthwt.grams ~ race + mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2131.57  -488.02    -1.16   521.87  1757.07 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2949.979    255.352  11.553   <2e-16 ***
## raceblack   -365.715    160.636  -2.277   0.0240 *  
## raceother   -285.466    115.531  -2.471   0.0144 *  
## mother.age     6.288     10.073   0.624   0.5332    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 715.7 on 185 degrees of freedom
## Multiple R-squared:  0.05217,    Adjusted R-squared:  0.0368 
## F-statistic: 3.394 on 3 and 185 DF,  p-value: 0.01909

Note that there are two coefficients estimated for the race variable (raceother and racewhite). What’s happening here?

When you put a factor variable into a regression, you’re allowing a different intercept at every level of the factor. In the present example, you’re saying that you want to model birthwt.grams as

Baby’s birthweight = Intercept(based on mother’s race) + β * mother’s age

We can rewrite this more succinctly as: \[ y=\mathrm{Intercept}_{race}+\beta\times age \] Essentially you’re saying that your data is broken down into 3 racial groups, and you want to model your data as having the same slope governing how birthweight changes with mother’s age, but potentially different intercepts. Here’s a picture of what’s happening.

# Calculate race-specific intercepts
intercepts <- c(coef(birthwt.lm)["(Intercept)"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceblack"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceother"])

lines.df <- data.frame(intercepts = intercepts,
                       slopes = rep(coef(birthwt.lm)["mother.age"], 3),
                       race = levels(birthwt$race))

ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +
  geom_point() +
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)

How do we interpret the 2 race coefficients? For categorical variables, the interpretation is relative to the given baseline. The baseline is just whatever level comes first (here, “white”). E.g., the estimate of raceother means that the estimated intercept is -285.47 higher among “other” race mothers compared to black mothers. Similarly, the estimated intercept is -365.72 higher for black mothers than white mothers.

Another way of putting it: Among mothers of the same age, babies of black mothers are born on average weighing -365.72g more than babies of white mothers.

Why is one of the levels missing in the regression?

As you’ve already noticed, there is no coefficient called “racewhite” in the estimated model. This is because this coefficient gets absorbed into the overall (Intercept) term.

Let’s peek under the hood. Using the model.matrix() function on our linear model object, we can get the data matrix that underlies our regression. Here are the first 20 rows.

head(model.matrix(birthwt.lm), 20)

##    (Intercept) raceblack raceother mother.age
## 1            1         1         0         19
## 2            1         0         1         33
## 3            1         0         0         20
## 4            1         0         0         21
## 5            1         0         0         18
## 6            1         0         1         21
## 7            1         0         0         22
## 8            1         0         1         17
## 9            1         0         0         29
## 10           1         0         0         26
## 11           1         0         1         19
## 12           1         0         1         19
## 13           1         0         1         22
## 14           1         0         1         30
## 15           1         0         0         18
## 16           1         0         0         18
## 17           1         1         0         15
## 18           1         0         0         25
## 19           1         0         1         20
## 20           1         0         0         28

Even though we think of the regression birthwt.grams ~ race + mother.age as being a regression on two variables (and an intercept), it’s actually a regression on 3 variables (and an intercept). This is because the race variable gets represented as two dummy variables: one for race == other and the other for race == black.

Why isn’t there a column for representing the indicator of race == white? This gets back to our colinearity issue. By definition, we have that

racewhite + raceother + raceblack = 1 = (Intercept)

This is because for every observation, one and only one of the race dummy variables will equal 1. Thus the group of 4 variables {racewhite, raceother, raceblack, (Intercept)} is perfectly colinear, and we can’t include all 4 of them in the model. The default behavior in R is to remove the dummy corresponding to the first level of the factor (here, racewhite), and to keep the rest.

Interaction terms

Let’s go back to the regression line plot we generated above.

ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +
  geom_point() +
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)

We can have similar plots by using the geom_smooth or stat_smooth commands in ggplot. Compare the plot above to the following.

ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +
  geom_point() +
  stat_smooth(method = "lm", se = FALSE, fullrange = TRUE)

In this case we have not only race-specific intercepts, but also race-specific slopes. The plot above corresponds to the model:

Baby’s birthweight = Intercept(based on mother’s race) + β(based on mother’s race) * mother’s age

We can rewrite this more succinctly as: \[ y=\mathrm{Intercept}_{race}+\beta_{race}\times age \] To specify this interaction model in R, we use the following syntax

birthwt.lm.interact <- lm(birthwt.grams ~ race * mother.age, data = birthwt)
summary(birthwt.lm.interact)

## 
## Call:
## lm(formula = birthwt.grams ~ race * mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2182.35  -474.23    13.48   523.86  1496.51 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)           2583.54     321.52   8.035 1.11e-13 ***
## raceblack             1022.79     694.21   1.473   0.1424    
## raceother              326.05     545.30   0.598   0.5506    
## mother.age              21.37      12.89   1.658   0.0991 .  
## raceblack:mother.age   -62.54      30.67  -2.039   0.0429 *  
## raceother:mother.age   -26.03      23.20  -1.122   0.2633    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 710.7 on 183 degrees of freedom
## Multiple R-squared:  0.07541,    Adjusted R-squared:  0.05015 
## F-statistic: 2.985 on 5 and 183 DF,  p-value: 0.01291

We now have new terms appearing. Terms like raceblack:mother.age are deviations from the baseline slope (the coefficient of mother.age in the model) in the same way that terms like raceblack are deviations from the baseline intercept. This models says that:

On average among white mothers, every additional year of age is associated with a 21.37g increase in the birthweight of the baby.

To get the slope for black mothers, we need to add the interaction term to the baseline. \[ \beta_{\text{raceblack}}=\beta_{\text{racewhite}}+\beta_{\text{raceblack:mother.age}}=\text{mother.age}+\text{raceblack:mother.age}=21.3727574+(-62.5379308)= -41.1651733 \] This slope estimate is negative, which agrees with the regression plot above.

Is a categorical variable in a regression statistically significant?

Last class we considered modelling birthweight as a linear function of mother’s age, allowing for a race-specific intercept for each of the three race categories. This model is fit again below.

birthwt.lm <- lm(birthwt.grams ~ race + mother.age, data = birthwt)

Here’s a visualization of the model fit that we wound up with. Note that while there are 3 lines shown, this is a visualization of just one model: birthwt.grams ~ race + mother.age. This model produces 3 lines because the coefficients of the race variable result in different intercepts.

# Calculate race-specific intercepts
intercepts <- c(coef(birthwt.lm)["(Intercept)"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceblack"],
                coef(birthwt.lm)["(Intercept)"] + coef(birthwt.lm)["raceother"])

lines.df <- data.frame(intercepts = intercepts,
                       slopes = rep(coef(birthwt.lm)["mother.age"], 3),
                       race = levels(birthwt$race))

ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +   
  geom_point() +
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)

At this stage we may be interested in assessing whether the race variable is statistically significant. i.e., Does including the race variable significantly improve the fit of our model, or is the simpler model birthwt.grams ~ mother.age just as good?

Essentially, we want to know if the race-specific intercepts capture significantly more variation in the outcome (birthweight) than the single intercept model, or if allowing for different intercepts isn’t doing much more than capturing random fluctuations in the data.

Here’s a picture of the two models we’re comparing:

library(gridExtra)
plot.complex <- ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +
  geom_point() +
  geom_abline(aes(intercept = intercepts, 
                  slope = slopes, 
                  color = race), data = lines.df)

# Single intercept model (birthwt.grams ~ mother.age)
p <- ggplot(birthwt, aes(x = mother.age, y = birthwt.grams))
plot.simple <- p + geom_point(aes(colour = race)) + stat_smooth(method = "lm")

grid.arrange(plot.complex, plot.simple, ncol = 2)

To test this hypothesis, we use the anova function (not to be confused with the aov function). This function compares two nested models, accounting for their residual sums of squares (how well they fit the data) and their complexity (how many more variables are in the larger model) to assess statistical significance.

# Fit the simpler model with mother.age as the only predictor
birthwt.lm.simple <- lm(birthwt.grams ~ mother.age, data = birthwt)

# Compare to more complex model
anova(birthwt.lm.simple, birthwt.lm)

## Analysis of Variance Table
## 
## Model 1: birthwt.grams ~ mother.age
## Model 2: birthwt.grams ~ race + mother.age
##   Res.Df      RSS Df Sum of Sq      F  Pr(>F)  
## 1    187 99154173                              
## 2    185 94754346  2   4399826 4.2951 0.01502 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

This output tells us that the race variable is statistically significant: It is unlikely that the improvement in fit when the add the race variable is simply due to random fluctuations in the data. Thus it is important to consider race when modeling how birthweight depends on the mother’s age.

Is an interaction term significant?

Assessing significance of interaction terms operates on the same principle. We once again ask whether the improvement in model fit is worth the increased complexity of our model. For instance, consider the example we saw last class, where we allowed for a race-specific slope in addition to the race-specific intercept from before.

birthwt.lm.interact <- lm(birthwt.grams ~ race * mother.age, data = birthwt)
summary(birthwt.lm.interact)

## 
## Call:
## lm(formula = birthwt.grams ~ race * mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2182.35  -474.23    13.48   523.86  1496.51 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)           2583.54     321.52   8.035 1.11e-13 ***
## raceblack             1022.79     694.21   1.473   0.1424    
## raceother              326.05     545.30   0.598   0.5506    
## mother.age              21.37      12.89   1.658   0.0991 .  
## raceblack:mother.age   -62.54      30.67  -2.039   0.0429 *  
## raceother:mother.age   -26.03      23.20  -1.122   0.2633    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 710.7 on 183 degrees of freedom
## Multiple R-squared:  0.07541,    Adjusted R-squared:  0.05015 
## F-statistic: 2.985 on 5 and 183 DF,  p-value: 0.01291

Here’s a side-by-side visual comparison of therace + mother.age model and the race + mother.age + race*mother.age interaction model.

plot.interact <- ggplot(mapping = aes(x = mother.age, y = birthwt.grams, color = race), data = birthwt) +   
  geom_point() +
  stat_smooth(method = "lm", se = FALSE, fullrange = TRUE)
grid.arrange(plot.interact, plot.complex, ncol = 2)

So, do the lines with different slopes fit the data significantly better than the common slope model? Let’s compare the two with the anova() function.

anova(birthwt.lm, birthwt.lm.interact)

## Analysis of Variance Table
## 
## Model 1: birthwt.grams ~ race + mother.age
## Model 2: birthwt.grams ~ race * mother.age
##   Res.Df      RSS Df Sum of Sq      F Pr(>F)
## 1    185 94754346                           
## 2    183 92431148  2   2323199 2.2998 0.1032

This \(p\)-value turns out to not be statistically significant. So even though the estimated slopes in the interaction model look very different, our estimates are quite variable, so we don’t have enough evidence to conclude that the interaction term (different slopes) is providing significant additional explanatory power over the simpler race + mother.age model.

Is my complex model signficantly better than a simpler one?

The testing strategy above applies to any two nested models. Here’s an example where we add in a few more variables and see how it compares to the race + mother.age model from earlier.

birthwt.lm.complex <- lm(birthwt.grams ~ mother.smokes + physician.visits + race + mother.age, data = birthwt)

summary(birthwt.lm.complex)

## 
## Call:
## lm(formula = birthwt.grams ~ mother.smokes + physician.visits + 
##     race + mother.age, data = birthwt)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2335.06  -455.16    31.74   499.29  1623.57 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)      3282.407    261.326  12.561  < 2e-16 ***
## mother.smokesyes -424.651    110.371  -3.847 0.000165 ***
## physician.visits   14.391     48.953   0.294 0.769102    
## raceblack        -444.340    156.586  -2.838 0.005057 ** 
## raceother        -445.161    119.666  -3.720 0.000265 ***
## mother.age          1.547      9.996   0.155 0.877155    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 691.7 on 183 degrees of freedom
## Multiple R-squared:  0.1241, Adjusted R-squared:  0.1001 
## F-statistic: 5.184 on 5 and 183 DF,  p-value: 0.000179

Let’s compare to our earlier model:

anova(birthwt.lm, birthwt.lm.complex)

## Analysis of Variance Table
## 
## Model 1: birthwt.grams ~ race + mother.age
## Model 2: birthwt.grams ~ mother.smokes + physician.visits + race + mother.age
##   Res.Df      RSS Df Sum of Sq      F    Pr(>F)    
## 1    185 94754346                                  
## 2    183 87567280  2   7187067 7.5098 0.0007336 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Highly significant! This is probably due to the fact that mother’s smoking status has a tremendously high association with birthweight.

Analysis of Covariance

ANCOVA in R, or Analysis of Covariance, is a powerful statistical method that combines aspects of analysis of variance (ANOVA) and regression analysis. It is employed to assess group mean differences while taking into account the influence of continuous covariates. ANCOVA is particularly valuable when researchers need to control for variables that might confound the interpretation of their results. This technique allows us to determine whether there are significant differences in the means of dependent variables among various groups, even when considering the potential impact of covariates.

The main difference between ANOVA (Analysis of Variance) and ANCOVA (Analysis of Covariance) lies in how they handle and incorporate variables, as well as their primary objectives:

1. Purpose

ANOVA: Compares the means of two or more groups to determine if there are statistically significant differences between them.

ANCOVA: Compares the means of two or more groups while controlling for the effects of one or more continuous covariates (i.e., control variables).

2. Variables Involved

ANOVA:

Independent variables (factors): Categorical variables that define the groups.

Dependent variable: Continuous outcome variable being compared across groups.

ANCOVA:

Independent variables (factors): Categorical variables that define the groups.

Dependent variable: Continuous outcome variable being compared across groups.

Covariates: Continuous variables that are included to control for their effects, improving the accuracy of group comparisons.

3. Example Scenarios

ANOVA: You want to test if different teaching methods (Group A, B, C) lead to different average test scores.

ANCOVA: You want to test if different teaching methods (Group A, B, C) lead to different average test scores while controlling for students’ prior knowledge (e.g., pre-test scores).

Mathematical Model

ANOVA

The ANOVA model can be expressed as:

\[ Y_{ij} = \mu + \alpha_i + \epsilon_{ij} \]

Where: - \(Y_{ij}\): Dependent variable (response)

• \(\mu\): Overall mean

• \(\alpha_i\): Effect of group \(i\)

• \(\epsilon_{ij}\): Random error

ANCOVA

The ANCOVA model adds a covariate to the equation:

\[ Y_{ij} = \mu + \alpha_i + \beta X_{ij} + \epsilon_{ij} \]

Where: - \(X_{ij}\): Covariate

• \(\beta\): Regression coefficient for the covariate

• Other terms (\(Y_{ij}\), \(\mu\), \(\alpha_i\), and \(\epsilon_{ij}\)) are as defined in the ANOVA model.

Assumptions

ANCOVA makes several assumptions about the data, such as:

Linearity between the covariate and the outcome variable at each level of the grouping variable.

This can be checked by creating a grouped scatter plot of the covariate and the outcome variable.

Homogeneity of regression slopes.

The slopes of the regression lines, formed by the covariate and the outcome variable, should be the same for each group. This assumption evaluates that there is no interaction between the outcome and the covariate. The plotted regression lines by groups should be parallel.

The outcome variable should be approximately normally distributed.
This can be checked using the Shapiro-Wilk test of normality on the model residuals.

Homoscedasticity or homogeneity of residuals variance for all groups.
The residuals are assumed to have a constant variance (homoscedasticity)

No significant outliers in the groups

Related libraries

library(tidyverse)
library(ggpubr)
library(rstatix)
library(broom)
library(datarium)

One-way ANCOVA

#make this example reproducible 
set.seed(10)

#create dataset
data <- data.frame(technique = rep(c("A", "B", "C"), each = 30),
                   current_grade = runif(90, 65, 95),
                   exam = c(runif(30, 80, 95), runif(30, 70, 95), runif(30, 70, 90)))

head(data)

##   technique current_grade     exam
## 1         A      80.22435 87.32759
## 2         A      74.20306 90.67114
## 3         A      77.80723 88.87902
## 4         A      85.79306 87.75735
## 5         A      67.55408 85.72442
## 6         A      71.76310 92.52167

Explore the data

Before we fit the ANCOVA model, we should first explore the data to gain a better understanding of it and verify that there aren’t any extreme outliers that could skew the results.

First, we can view a summary of each variable in the dataset:

summary(data)

##   technique         current_grade        exam      
##  Length:90          Min.   :65.43   Min.   :71.17  
##  Class :character   1st Qu.:71.79   1st Qu.:77.27  
##  Mode  :character   Median :77.84   Median :84.69  
##                     Mean   :78.15   Mean   :83.38  
##                     3rd Qu.:83.65   3rd Qu.:89.22  
##                     Max.   :93.84   Max.   :94.76

Next, we can use the dplyr package to easily find the mean and the standard deviation of both the current grades and the exam scores for each studying technique:

#load dplyr
library(dplyr)

data %>%
  group_by(technique) %>%
  summarise(mean_grade = mean(current_grade),
            sd_grade = sd(current_grade),
            mean_exam = mean(exam),
            sd_exam = sd(exam))

## # A tibble: 3 × 5
##   technique mean_grade sd_grade mean_exam sd_exam
##   <chr>          <dbl>    <dbl>     <dbl>   <dbl>
## 1 A               79.0     7.00      88.5    3.88
## 2 B               78.5     8.33      81.8    7.62
## 3 C               76.9     8.24      79.9    5.71

Please plot a boxplot showing the relationship between exam grades with technique

# ?boxplot

Similarly, we can also use boxplots to visualize the distribution of current grades based on studying technique:

# ?boxplot

Next, we’ll fit the ANCOVA model using exam score as the response variable, studying technique as the predictor (or “treatment”) variable, and current grade as the covariate. We start with the complete model, and consider the interaction between variables.

We’ll use the Anova() function in the car package to do so, just so we can specify that we’d like to use type III sum of squares for the model, since type I sum of squares is dependent upon the order that the predictors are entered into the model

library(car)
# ?lm
# ?Anova

Now let’s fit a simpler model without interaction.

# ?lm

Post Hoc Tests

Although the ANCOVA results told us that studying technique had a statistically significant effect on exam scores, we need to run post hoc tests to actually find out which studying techniques differ from each other.

To do so, we can use the glht() function within the multcomp package in R to perform Tukey’s Test for multiple comparisons:

library(multcomp)
#fit the ANCOVA model

# notice here you need to make your technique as a factor variable first, otherwise you will notice an error
data$technique <- as.factor(data$technique)
 
# ?glht
# for the glht function you will use, set: linfct = mcp(technique = "Tukey")

# after fitting the General linear hypothese, use summary to take a look at the result 

What if you would like to take a look at the confidence interval of the mean difference using different teaching techniques? Use function ‘confint()’ here.

#view the confidence intervals associated with the multiple comparisons
# confint() one the glht hypotheses you just did